Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{8n + 16}{n^2 + 11n + 18} \times \dfrac{n^2 + 9n}{n + 10} $
Explanation: First factor the quadratic. $p = \dfrac{8n + 16}{(n + 9)(n + 2)} \times \dfrac{n^2 + 9n}{n + 10} $ Then factor out any other terms. $p = \dfrac{8(n + 2)}{(n + 9)(n + 2)} \times \dfrac{n(n + 9)}{n + 10} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 8(n + 2) \times n(n + 9) } { (n + 9)(n + 2) \times (n + 10) } $ $p = \dfrac{ 8n(n + 2)(n + 9)}{ (n + 9)(n + 2)(n + 10)} $ Notice that $(n + 2)$ and $(n + 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 8n(n + 2)\cancel{(n + 9)}}{ \cancel{(n + 9)}(n + 2)(n + 10)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $p = \dfrac{ 8n\cancel{(n + 2)}\cancel{(n + 9)}}{ \cancel{(n + 9)}\cancel{(n + 2)}(n + 10)} $ We are dividing by $n + 2$ , so $n + 2 \neq 0$ Therefore, $n \neq -2$ $p = \dfrac{8n}{n + 10} ; \space n \neq -9 ; \space n \neq -2 $